The sides of a triangle are $a^{2} - 3 a b + 8$ , $4 a^{2} + 5 a b - 3$ and $6 - 3 a^{2} + 4 a b$ . Then its perimeter is

$2 a^{2} + 6 a b + 11$

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$a^{2} + 2 a b + 11$

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$2 a^{2} + a b + 11$

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$2 a^{2} + 2 a b$

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Solution

The correct option is A
$2 a^{2} + 6 a b + 11$

Sum of the three sides gives the perimeter of the triangle.

i.e. Perimeter = $(a^{2} - 3 a b + 8) + (4 a^{2} + 5 a b - 3) + (6 - 3 a^{2} + 4 a b)$

= $a^{2} + 4 a^{2} - 3 a^{2} - 3 a b + 5 a b + 4 a b + 8 - 3 + 6$

= $2 a^{2} + 6 a b + 11$

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Similar questions

$If the sides of a triangle are a^{2} - 3 a b + 8 units, 4 a^{2} + 5 a b - 3 units and 6 - 3 a^{2} + 4 a b units, then its perimeter is ____ units.$

If the sides of a triangle are a^{2} - 3 a b + 8 units, 4 a^{2} + 5 a b - 3 units and 6 - 3 a^{2} + 4 a b units, then find its perimeter.

State True or False:

On subtracting $a^{2} + a b + b^{2}$ from $4 a^{2} - 3 a b + 2 b^{2}$ , the answer is $3 a^{2} - 4 a b + b^{2}$ .

14 a^{2} + 20 a + 13

3 a^{2} + 5 a + 1

a^{2} + 10 a - 6

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